In the figure, two circle intersect at P and Q. O is the centre of the smaller circle which lies on the circumference of the larger circle and RO is joined and produced to meet QS at X. Prove that: RX ⊥ QS, QX = SX, RQ = RS.
Answers
Answer:
Explanation:
Construction: Join OP and OQ
1) chord OP = chord OQ [Radii of circle]
2) arc OP = arc OQ [from statement 1, equal arcs subtended from equal
chords]
3) <QRX = <SRX [Equal inscribed angles subtended from equal arcs]
4) <POQ = 2 <PSQ [Relation of inscribed and central angles]
5) <POQ + <QRS = 180 [Sum of opposite angles of cyclic quad is
supplementary]
6) 2<PSQ + <QRS = 180 [From statement 4 and 5]
or, <QRS = 180 - 2<PSQ
7) <QRS + <PSQ + <RQS = 180 [Sum of angles of triangle RQS]
8) 180 - 2<PSQ + <PSQ + <RQS = 180 [From statement 6 and 7]
9) <PSQ = <RQS [from statement 7]
10) RQS is isosceles triangle [from statement 9, base angles of isosceles
triangle is equal]
11) In triangles RQX and RSX,
i) <QRX = <XRS [A] [from statement 3]
ii) RQ = RS [S] [from statement 10, base sides of isosceles triangle
is equal]
iii) <RQX = <RSX [A] [From statement 9]
12) triangle RQX ≅ triangle RSX [from statement 11 by A.S.A axiom]
13) RQ = RS [Corresponding sides of congruent triangles]
14) QX = SX [Same as 13]
15) <RXS = <RXQ [Corresponding angles of congruent triangles]
16) <RXS + <RXQ = 180 [Sum of angles in st. line]
17) <RXS + <RXS = 180 [from statement 15 and 16]
or, 2<RXS = 180
18) <RXS = 90
19) RX ⊥ QS
PROVED